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Question
On what intervals of t is the curve described by the given parametric equation concave up? Concave down?
x=t^2; y=t^(3) + 3t

Answer
The signage of the second order derivative( ie positive, negative or zero) would help to discover the nature of concavity for the curve at various intervals.

dy/dx =dy/dt * dt/dx  = (3t^2 + 3)/ (2t)

(d^2 y)/(dx^2)= d/dx (dy/dx) = d/dt (dy/dx) * ( dt/dx)

         =  [2t*(6t)-2(3t^2 + 3)]/(4t^2)  * 1/(2t)

         =  (6t^2 - 6)/(8t^3) =  (3t^2 - 3)/(4t^3)      

(Note that since dy/dx is actually a function in t, implicit differentiation must be employed when seeking the expression for (d^2 y)/(dx^2) )

When the curve is concave downwards,  (d^2 y)/(dx^2)<0 =====> (3t^2 - 3)/(4t^3) < 0
         [(t-1)(t+1)]/ (t^3) < 0

         ie  t < -1   or   0 < t < 1  (shown)    


When the curve is concave downwards,  (d^2 y)/(dx^2)   =====> (3t^2 - 3)/(4t^3) > 0
         [(t-1)(t+1)]/ (t^3) > 0

         ie  -1 < t < 0  or t > 1  (shown)  

Hope this helps. Peace.  

Calculus

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